Let $$H: \frac{-x^2}{a^2}+\frac{y^2}{b^2}=1$$ be the hyperbola, whose eccentricity is $$\sqrt{3}$$ and the length of the latus rectum is $$4 \sqrt{3}$$. Suppose the point $$(\alpha, 6), \alpha>0$$ lies on $$H$$. If $$\beta$$ is the product of the focal distances of the point $$(\alpha, 6)$$, then $$\alpha^2+\beta$$ is equal to

Consider a hyperbola $$\mathrm{H}$$ having centre at the origin and foci on the $$\mathrm{x}$$-axis. Let $$\mathrm{C}_1$$ be the circle touching the hyperbola $$\mathrm{H}$$ and having the centre at the origin. Let $$\mathrm{C}_2$$ be the circle touching the hyperbola $$\mathrm{H}$$ at its vertex and having the centre at one of its foci. If areas (in sq units) of $$C_1$$ and $$C_2$$ are $$36 \pi$$ and $$4 \pi$$, respectively, then the length (in units) of latus rectum of $$\mathrm{H}$$ is

If the foci of a hyperbola are same as that of the ellipse $$\frac{x^2}{9}+\frac{y^2}{25}=1$$ and the eccentricity of the hyperbola is $$\frac{15}{8}$$ times the eccentricity of the ellipse, then the smaller focal distance of the point $$\left(\sqrt{2}, \frac{14}{3} \sqrt{\frac{2}{5}}\right)$$ on the hyperbola, is equal to